Q82 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics

$\int \frac{1}{7-6 x-x^2} d x=$

  • A
    $\frac{1}{4} \log \left(\frac{7+x}{1-x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • Correct answer
    B
    $\frac{1}{8} \log \left(\frac{7+x}{1-x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • C
    $\frac{1}{16} \log \left(\frac{7+x}{1-x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • D
    $\frac{1}{32} \log \left(\frac{7+x}{1-x}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

Let $\begin{aligned} \mathbf{I} & =\int \frac{1}{7-6 x-x^2} \mathrm{~d} x \\ & =\int \frac{1}{7-6 x-x^2-9+9} \mathrm{~d} x \\ & =\int \frac{1}{16-\left(x^2+6 x+9\right)} \mathrm{d} x \\ & =\int \frac{1}{(4)^2-(x+3)^2} \mathrm{~d} x \\ & =\frac{1}{8} \log \left|\frac{4+x+3}{4-(x+3)}\right|+\mathrm{c} \\ & =\frac{1}{8} \log \left|\frac{7+x}{1-x}\right|+\mathrm{c} \end{aligned}$

Hence, the correct answer is option: 'B'.

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